**1.**√x(√9-4/3)=√25

The sqrt is just over the 9.

**2.**(4-3*√2)

^{2}

**3.**√135b

^{2}c

^{3}d*√5b

^{2}d

The sqrt is over 135b

^{2}c^{3}d and 5b^{2}d The sqrt is just over the 9.

The sqrt is over 135b^{2}c^{3}d and 5b^{2}d

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Saugus, MA

1) √x(√9-[4/3])=√25

√x(3-[4/3])=5

√x(5/3)=5

√x(5/3)(3/5)=5(3/5)

√x=3

x=9

3) (√135b^{2}c^{3}d)*(√5b^{2}d)

3bc(√15cd)*b(√5d)

3bc(√5d√3c)*b(√5d)

(3bc)(b)(√5d)(√5d)(√3c)

(3b^{2}c)(5d)(√3c)

15b^{2}cd√3c

Arthur:

4th line:

√(X . 5/3) = 5

then:

X(5/3) = 25

X = 25. 3/5 = 15

You forgot to square the 5 ( right side).

Parviz,

I thought the first √ was only for x, not for what is in the parentheses. This problem is misleading because we don't know for sure what the radicand is for that very first radical sign. Your first comment says it all; this makes me think you too aren't really sure what is under that first radical sign:the x only or the x and everything in parentheses. Also, the way I solved it, I multiplied 5 by 3/5 to get 3 and then squared the 3.

I guess the only one who knows for sure is the student and so far there is no feedback. We'll just have to keep trying to figure out exactly what the problem is and take it from there.

Arthur D.

Yes, Student's writing is not very clear. They should inscribe radical's argument inside parenthesis.

Woodland Hills, CA

Square both sides of the equation:

X ( 3 - 4/3 ) = 25

X = 25 / ( 3 - 4/3 )

X = 25 /( 5/3)

X = 75/5 = 15

2.

( 4 - 3 √2 ) = 16 - 24 √2 + 18 = 34 - 24 √2 ( Use of the relation of ( a ±b)^{2} = a^{2} + b^{2
}±2ab

3.

√( 135b^{2} c^{3} d * √(5b^{2} d ) =

3bc √(15 cd ) * b √5d ) =

15b^{2} d c^{2}√(5c)

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